proof of existence of the essential supremum
Suppose that (Ω,ℱ,μ) is a σ-finite measure space and 𝒮 is a collection of measurable functions
f:Ω→ˉℝ. We show that the essential supremum
of 𝒮 exists and furthermore, if it is nonempty then there is a sequence fn∈𝒮 such that
As any -finite measure is equivalent
to a probability measure (http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure), we may suppose without loss of generality that is a probability measure. Also, without loss of generality, suppose that is nonempty, and let consist of the collection of maximums of finite sequences of functions in . Then choose any continuous
and strictly increasing . For example, we can take
As is a bounded and measurable function for all , we can set
Then choose a sequence in such that . By replacing by the maximum of if necessary, we may assume that for each . Set
Also, every is the maximum of a finite sequence of functions in . Therefore, there exists a sequence such that
Then,
It only remains to be shown that is indeed the essential supremum of . First, by continuity of and the dominated convergence theorem,
Similarly, for any ,
It follows that is a nonnegative function with nonpositive integral, and so is equal to zero -almost everywhere. As is strictly increasing, and therefore -almost everywhere.
Title | proof of existence of the essential supremum |
---|---|
Canonical name | ProofOfExistenceOfTheEssentialSupremum |
Date of creation | 2013-03-22 18:39:25 |
Last modified on | 2013-03-22 18:39:25 |
Owner | gel (22282) |
Last modified by | gel (22282) |
Numerical id | 7 |
Author | gel (22282) |
Entry type | Proof |
Classification | msc 28A20 |